Looking for something else?

Not the answer you are looking for? Search for more explanations.

- billyjean

How do you measure the distance between a point and a line for a circle.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- billyjean

How do you measure the distance between a point and a line for a circle.

- katieb

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- Owlcoffee

Well, Let's take a look at the scenario, shall we?.
What we will do is take generalized scenario, with a general line and a general point in the plane.
We can begin by stating that we will use an ortonormated reference point, this does mean that the axis will be perpendicular to each other, like the one you are usd to.
So, I will explain as much as I can, because this is a little confusing.
Let us be given the following:
1) r) ax+by+c=0 and M(xo,yo).
Let the segment MH be perpendicular to line (r) being "H" a point that belongs to the line:
|dw:1433974890346:dw|
(I'll further explain the rest of the drawing as I go).
With this given information, we can begin to analyze,
now, the distance from point "M" to line "r" will be defined by the distance from point M to point H:
\[dist(M,r)=dist(M,H)=\left| \left| \vec {MH} \right| \right|\]
This means that the distance between those two points will be equal to the measure of the vector that is defined by M and H, so we will limit us to finding the module of the vector MH.
Let a vector whose tail resides at the origin be:
\[\vec u = [a,b], \vec u \perp (r) \]
We just created a vector, to help us out, and this vector is perpendicular to the line, and has coordinates [a,b].
This will mean that the vector MH and that vector u are colineal, and since they both have an angle of 0:
\[\left| \vec{MH} \times \vec u \right|=\left| \left| \vec{MH} \right| \right|.\left| \left| \vec u \right| \right|\]
but, now we will determine the coordinates of the vector MH and u:
\[\vec u = [a,b]\]
\[\vec {MH}=[(x_o - x_h),(y_o-y_h)]\]
being xh and yh the coordinates of the point "H".
So, therefore:
\[\left| \vec {MH} \times \vec u \right|=[a(x_o- x_h)+b(x_o - y_h)]\]
but, we can determine the module of vector u:
\[\vec u = \sqrt{a^2+b^2}\]
and substituting:
\[\left| \left| \vec {MH} \right| \right|.\sqrt{a^2+b^2}=[a(x_o-x_h)+b(y_o-y_h)]\]
We will do the distributive and order it up:
\[\left| ax_o+by_o-ax_h-by_h \right|=\left| \left| \vec {MH} \right| \right|.\sqrt{a^2+b^2}\]
but, H is a point belonging to the line (r) so therefore:
\[ax_h+by_h+c=0\]
and we'll isolate "c":
\[c=-ax_h-by_h\]
this appears in the equality, so we can just substitute it:
\[\left| ax_o+by_o+c \right|=\left| \left| \vec {MH} \right| \right|.\sqrt{a^2+b^2}\]
But we defined in the beginning that the module od the vector MH represents the distance from the point to the line:
\[\left| ax_o+by_o+c \right|=dist(M,r).\sqrt{a^2+b^2}\]
so therefore:
\[dist(M,r)=\frac{ ax_o+by_o+c }{ \sqrt{a^2+b^2} }\]
And that is the formula that can allow us to calculate the distance from a point, to a line.

Looking for something else?

Not the answer you are looking for? Search for more explanations.